$ E = \left[\begin{array}{rr}2 & 3 \\ 4 & -2\end{array}\right]$ $ A = \left[\begin{array}{rr}-2 & 2 \\ -1 & -1\end{array}\right]$ What is $ E A$ ?
Because $ E$ has dimensions $(2\times2)$ and $ A$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ E A = \left[\begin{array}{rr}{2} & {3} \\ {4} & {-2}\end{array}\right] \left[\begin{array}{rr}{-2} & \color{#DF0030}{2} \\ {-1} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{2}\cdot{-2}+{3}\cdot{-1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{-2}+{3}\cdot{-1} & ? \\ {4}\cdot{-2}+{-2}\cdot{-1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{-2}+{3}\cdot{-1} & {2}\cdot\color{#DF0030}{2}+{3}\cdot\color{#DF0030}{-1} \\ {4}\cdot{-2}+{-2}\cdot{-1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{2}\cdot{-2}+{3}\cdot{-1} & {2}\cdot\color{#DF0030}{2}+{3}\cdot\color{#DF0030}{-1} \\ {4}\cdot{-2}+{-2}\cdot{-1} & {4}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-7 & 1 \\ -6 & 10\end{array}\right] $